Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{q^2 - 5q + 6}{-5q + 45} \div \dfrac{q - 2}{-3q + 27} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{q^2 - 5q + 6}{-5q + 45} \times \dfrac{-3q + 27}{q - 2} $ First factor the quadratic. $p = \dfrac{(q - 2)(q - 3)}{-5q + 45} \times \dfrac{-3q + 27}{q - 2} $ Then factor out any other terms. $p = \dfrac{(q - 2)(q - 3)}{-5(q - 9)} \times \dfrac{-3(q - 9)}{q - 2} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ (q - 2)(q - 3) \times -3(q - 9) } { -5(q - 9) \times (q - 2) } $ $p = \dfrac{ -3(q - 2)(q - 3)(q - 9)}{ -5(q - 9)(q - 2)} $ Notice that $(q - 9)$ and $(q - 2)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -3\cancel{(q - 2)}(q - 3)(q - 9)}{ -5(q - 9)\cancel{(q - 2)}} $ We are dividing by $q - 2$ , so $q - 2 \neq 0$ Therefore, $q \neq 2$ $p = \dfrac{ -3\cancel{(q - 2)}(q - 3)\cancel{(q - 9)}}{ -5\cancel{(q - 9)}\cancel{(q - 2)}} $ We are dividing by $q - 9$ , so $q - 9 \neq 0$ Therefore, $q \neq 9$ $p = \dfrac{-3(q - 3)}{-5} $ $p = \dfrac{3(q - 3)}{5} ; \space q \neq 2 ; \space q \neq 9 $